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The implicit midpoint rule advances an autonomous ordinary differential equation byIt is a second-order, A-stable, time-symmetric numerical method.

The implicit midpoint rule is the midpoint method with $y_{n+1}=y_n+hf(t_n+h/2,(y_n+y_{n+1})/2)$. It is a one-stage implicit Runge-Kutta method.

For the Strang splittingreversing $t$ reverses all three factors, soIt is therefore a time-symmetric numerical method. Multiplication of the three exponential series shows agreement with $e^{t(A+B)}$ through degree two. Equivalently, the Baker--Campbell--Hausdorff formula gives an odd modified generatorfor a matrix $D$ made from nested commutators. Exponentiating givesfor a matrix $C$ depending on $A$ and $B$.

Let ${\Phi }_h$ be the one-step map of a numerical method. It is a time-symmetric numerical method when reversing the step exactly reverses the update:

Let ${\phi }_h$ be the exact flow map and suppose the method has order $p$ with a nonzero leading local truncation error:Inverting this expansion changes the sign of its leading perturbation, sowhere transport by the exact flow only changes $d$ by $O(h)$ and hence does not affect the leading parity. On the other hand, replacing $h$ by $-h$ in the first expansion givesTime symmetry equates these expressions, so $(-1{)}^{p+1}=-1$. Hence $p+1$ is odd and

A time-symmetric numerical method is a one-step numerical method with a locally invertible step map ${\Phi }_h$ satisfyingFor a nonautonomous ordinary differential equation, include time in the state or explicitly reverse the step's time endpoints. The definition means that applying the same rule backward undoes its forward step; it does not by itself claim small discretization error.

The implicit midpoint rule illustrates this directly: swapping $y_n,y_{n+1}$ and replacing $h$ by $-h$ leaves its midpoint equation unchanged, wherever the local solve is unique. By contrast, forward Euler method has factor $1+h\lambda$ on $y^{\prime }=\lambda y$, and the backward factor $1-h\lambda$ gives product $1-h^2{\lambda }^2$, not one.

Assume a consistent method has a smooth integer-power local-error expansion of finite order $p$, with nonzero leading defect $h^{p+1}E(x)$. Composing its forward and backward steps leaves leading defect $[1+(-1{)}^{p+1}]h^{p+1}E(x)$. Symmetry forces this coefficient to vanish. Thus $p+1$ is odd and $p$ is even. The finite leading-error and regular expansion assumptions are needed for this parity argument.